Question: $\vec w = \left(2,-\dfrac23\right)$ $6\vec w= ($
Answer: In general, the scalar multiple of $k$ times $\vec u$ is this: $k\vec u = k(u_x, u_y) = (ku_x, ku_y)$. So, here's how we find $6 \vec{w}$ : $\begin{aligned} {6}\vec w = {6} \cdot \left(2,-\dfrac23\right) &= \left({6} \cdot 2, {6} \cdot \left(-\dfrac23\right)\right) \\\\ &= (12,-4) \end{aligned}$ The answer is $ (12,-4) $.